# Transistor(BJT) Zenner diode voltage regulator circuit

calculate the regulated output voltage of a series voltage regulator circuit as shown in figure

Given $R_1$= 3.3 kOhm, $R_2$= 2.2 kOhm,$R_3$=5 kOhm,$R_4$=10 kOhm, $V_z$=10 V.

simulate this circuit – Schematic created using CircuitLab

Method followed in my book:

$$V_{R2}=V_{BE2}+V_Z=0.7+10=10.7 V$$ $$V_{R2}=V_o(\frac{R_2}{R_2+R_1})$$

But the second equation is valid only if current through zenner diode is 0. Please clear my doubt. If there is a way of evaluating this circuit without approximation then it will be highly helpful if you would post the solution.

• You calculated Vout=26.7V.
• output current is Vout/RL = 267mA
• Q1 dissipates 3.5W

For extra credit, you can prove the approximation you did is appropriate:

• Replace Q1 with a transistor which can actually handle this current and dissipation. 2N3904 cannot. I suggest D45H11, which will have hFe around 200.
• Calculate current through R4, with input voltage varying from 30 to 40V
• Notice circuit does not work, as Q1 base current is higher than what R4 can provide. Set R4 to proper value like 1k.
• Notice Q2 gets quite hot for Vin=40V. R4 would be much better as a current source, but let's leave it like that for now.
• Ic(Q1) is 1.2mA ... 11.3mA
• You get Ib(Q1)
• Calculate offset voltage at output due to Ib(Q1) applied to the impedance seen by Q1's base, which is R1//R2//R3 if we neglect zener impedance.

=> You get output voltage variation due to Q1 base current.

Compare to Zener accuracy and temperature drift from datasheet.

I find that output voltage variation due to Q1 base current is smaller than the zener 5% tolerance. Therefore, the approximation was legit.

• $$V_{CQ2}=V_{BEQ1}+V_o=26.7+0.7=27.4V$$ $$I_{R4}=\frac{40-27.4}{10000}=1.26mA$$ $$I_{R4}=\frac{40-27.4}{1000}=12.6mA$$ Sir I dont understant why you said the following statement(I cant understand the physics here) Notice circuit does not work, as Q1 base current is higher than what R4 can provide. And at this point have you calculated $I_{B1}$ by $I_{B1}=200I_{C1}$? – Soumee Mar 4 '17 at 14:39
• By "Notice Q2 gets quite hot for Vin=40V. R4 would be much better as a current source" do you mean that power dissipation in $Q_2$ will be very high? Also why would $R_4$ act as a current source? – Soumee Mar 4 '17 at 14:43
• I was pointing out a shortcoming of this particular circuit, which is that it only works well for a small range of input voltages. We have Vout=about 20V. If Vin is only 25V, R4 should be lowered to allow enough base current for Q1. But then, this value of R4 becomes too low for Vin=40V. A real implementation would replace R4 with a current source to solve this problem. But if your homework says Vin=40V you can ignore this. – peufeu Mar 4 '17 at 14:47

That is a reasonable approximation. The current flowing into the transistor base will be small, you cant calculate it anyway without more details about the transistors. The other error is the current through R3 which is Vbe2/R3 again small compared to the current through R1 and R2.

Vr2 is 10.7v. current through R3 or the zener is 0.7v/5k , or 0.14ma.

Current through r2 is 10.7v / 2.2k , or 5ma. Total current through r1 ISS 5.14ma. vooktage drop over r1 is 5.14ma x 3.3k or 16v.

So the. Output voltage Mut be 16v + 10.7v or 27v

• Understood. Sir, you are neglecting the base current through $Q_2$ by approximation – Soumee Mar 4 '17 at 14:21
• The whole point of that excercise is to show how small the current through the zener is vs. the current through r1 and r2. The collector current through q2 is about 1ma. so its base current is 20ua, to be conservative, and likely 5ua. It makes no practical difference. – dannyf Mar 4 '17 at 15:26
• i did a quick sim of the circuit: Vout = 26.4v, Ir1 = 4.8ma, Iz = 130ua, Ib2 (Q2's base current) = 3.3ua. So Ir1 >> Iz >> Ib2. – dannyf Mar 4 '17 at 18:24